Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.6 Inverse Trigonometric Functions - 6.6 Exercises - Page 481: 39

Answer

$\frac{\pi}{6}$

Work Step by Step

$$g'(x)=(\arcsin(\frac{x}{4})+x(\arcsin(\frac{x}{4}))')+\frac{-2x}{2\sqrt{16-x^2}}$$ $$g'(x)=(\arcsin(\frac{x}{4})+x(\frac{x}{4})'(\arcsin)'(\frac{x}{4}))+\frac{-2x}{2\sqrt{16-x^2}}$$ $$g'(x)=(\arcsin(\frac{x}{4})+\frac{x}{4}\frac{1}{\sqrt{1-(\frac{x}{4})^{2}}})+\frac{-2x}{2\sqrt{16-x^2}}$$ $$g'(x)=(\arcsin(\frac{x}{4})+\frac{x}{4}\frac{1}{\sqrt{1-(\frac{x}{4})^{2}}})-\frac{x}{\sqrt{16-x^2}}$$ so: $$g'(2)=(\arcsin(\frac{2}{4})+\frac{2}{4}\frac{1}{\sqrt{1-(\frac{2}{4})^{2}}})-\frac{2}{\sqrt{16-2^2}}=\frac{\pi}{6}$$
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